This is in continuation to Expected value of conditional events
In a cricket match, event A is a subset of all possible moves taken up by the fielding team. The batsman has a set of moves, and, batsman's strategy relies on the move taken up by the fielding team. Say, $B$ is the set of moves that defeats him (or, incurs some cost $\mathcal{C}$), i.e., $P(B/A)$ is known. $\mathcal{C}$ is a constant.
Let $A$ and $\bar{A}$ be the set of event chosen by the fielding side. And $B$ and $\bar{B}$ be the set of moves pursued by the batsman. Then all possible event combinations will be, $A \cap B$, $A \cap \bar{B}$, $\bar{A} \cap B$ and $\bar{A} \cap \bar{B}$. Given the problem, the associated probabilities calculated as, $P(B/A)P(A)$, $P(\bar{B}/A)P(A)$, $P(B/\bar{A})P(\bar{A})$ and $P(\bar{B}/\bar{A})P(\bar{A})$ respectively.
The batsman is willing to know the expected total cost s/he might incur if the fielding team chooses the strategy-set $A$.Batsman does not care about $\bar{A}$.
The expected cost incurred will be: $\mathcal{C}\left( 1 \times P(B/A)P(A) + 0 \times P(\bar{B}/A)P(A) + 0 \times P(B/\bar{A})P(\bar{A}) + 0 \times P(\bar{B}/\bar{A})P(\bar{A}) \right)$
Is my argument alright?